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A003440
Number of binary vectors with restricted repetitions.
(Formerly M2666)
6
1, 1, 3, 7, 17, 42, 104, 259, 648, 1627, 4098, 10350, 26202, 66471, 168939, 430071, 1096451, 2799072, 7154189, 18305485, 46885179, 120195301, 308393558, 791882862, 2034836222, 5232250537, 13462265079, 34657740889, 89272680921, 230069128392
OFFSET
0,3
COMMENTS
The sum of squared terms in row n of A104402 = 2*a(n) for n>0. - Paul D. Hanna, Mar 06 2005
From Jean-Pierre Levrel, Nov 26 2014: (Start)
The title "Binary Sequences with Restricted Repetitions," given the A003440 series, does not specify the type of restrictions used. After reading the article by K. A. Post, "Binary Sequences with Restricted Repetitions," it appears that the A003440 series corresponds to the following cases:
- Number of repetitions limited to two,
- Each sequence must begin with a zero.
It is important to consider these two hypotheses to interpret the series. I also think that the second constraint is not useful and could usefully be deleted. In this case, the series should be doubled from the second term and would become 1, 2, 6, 14, 34, 84, ..., i.e., A177790.
(End)
REFERENCES
N. J. A. Sloane and Simon Plouffe, The Encyclopedia of Integer Sequences, Academic Press, 1995 (includes this sequence).
LINKS
Jean-Luc Baril and José L. Ramírez, Fibonacci and Catalan paths in a wall, 2023.
K. A. Post, Binary Sequences with Restricted Repetitions, Report 74-WSK-02, Math. Dept., Tech. Univ. Eindhoven, May. 1974.
FORMULA
G.f.: {(1-x)^2 * sqrt[(1+x+x^2)/(1-3x+x^2)] + x^2 - 1}/(2x^2) (conjectured). - Ralf Stephan, Mar 28 2004
a(n) = Sum_{k=0..n} (C(k, n-k) + C(k+1, n-k-1))^2/2 for n>0, with a(0)=1. - Paul D. Hanna, Mar 06 2005
Conjecture: (n+2)*a(n) +3*(-n-1)*a(n-1) +(n-2)*a(n-2) +(-n+1)*a(n-3) +3*(n-4)*a(n-4) +(-n+5)*a(n-5)=0. - R. J. Mathar, Jun 07 2013
Recurrence: (n-2)*(n-1)*(n+2)*a(n) = 2*(n-2)*n*(n+1)*a(n-1) + (n-1)*(n^2 - 2*n - 4)*a(n-2) + 2*(n-3)*(n-2)*n*a(n-3) - (n-4)*(n-1)*n*a(n-4). - Vaclav Kotesovec, Feb 12 2014
a(n) ~ sqrt(6+14/sqrt(5)) * (3+sqrt(5))^n / (sqrt(Pi*n) * 2^(n+1)). - Vaclav Kotesovec, Feb 12 2014
Equivalently, a(n) ~ phi^(2*n + 2) / (5^(1/4) * sqrt(Pi*n)), where phi = A001622 is the golden ratio. - Vaclav Kotesovec, Dec 08 2021
MATHEMATICA
Flatten[{1, Table[Sum[(Binomial[k, n-k]+Binomial[k+1, n-k-1])^2/2, {k, 0, n}], {n, 1, 20}]}] (* Vaclav Kotesovec, Feb 12 2014 *)
a[r_, s_] /; r<0 || s<0 = 0; a[r_ /; 0 <= r <= 2, 0] = 1; a[r_ /; r>2, 0] = 0; a[0, s_ /; s >= 1] = 0; a[r_, s_] := a[r, s] = a[r-2, s-2] + a[r-2, s-1] + a[r-1, s-2] + a[r-1, s-1]; a[n_] := a[n, n]; Table[a[n], {n, 0, 29}] (* Jean-François Alcover, Jan 19 2015, after given recurrence *)
PROG
(PARI) {a(n)=polcoeff(((1-x)^2*sqrt((1+x+x^2)/(1-3*x+x^2))+x^2-1)/(2*x^2)+x*O(x^n), n)} \\ Paul D. Hanna, Mar 06 2005
(PARI) {a(n)=if(n==0, 1, sum(k=0, n, (binomial(k, n-k)+binomial(k+1, n-k-1))^2)/2)} \\ Paul D. Hanna, Mar 06 2005
CROSSREFS
KEYWORD
nonn
EXTENSIONS
Typo in second formula corrected by Vaclav Kotesovec, Feb 12 2014
More terms from Vincenzo Librandi, Feb 13 2014
STATUS
approved